We take advantage of the symmetry, indeed expand on it. Let
$$I=\int \frac{\sin^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta} \qquad\text{and}\qquad J=\int \frac{\cos^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta}.$$
Note that
$$\frac{\sin^3\theta}{\sin^3\theta-\cos^3\theta}=1+ \frac{\cos^3\theta}{\sin^3\theta-\cos^3\theta},$$
and therefore
$$I-J=\theta.$$
If we can find $I+J$ we will be finished.
So we want to find
$$\int\frac{\sin^3\theta+\cos^3\theta}{\sin^3\theta-\cos^3\theta}\,d\theta=
\int\frac{(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta)}{(\sin\theta-\cos\theta)(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta) }\,d\theta.$$
Let $u=\sin\theta-\cos\theta$. Then $du=(\cos\theta+\sin\theta)\,d\theta$. Also,
$u^2=1-2\sin\theta\cos\theta$. From this we find that $\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta=\frac{1+u^2}{2}$ and $\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta=\frac{3-u^2}{2}$. Thus
$$I+J=\int\frac{1+u^2}{u(3-u^2)}\,du.$$
We do a partial partial fraction decomposition:
$$\frac{1+u^2}{u(3-u^2)}=\frac{1}{3}\left(\frac{1}{u}+\frac{4u}{3-u^2}\right).$$
Integrate: $I+J=(1/3)\ln\left(\dfrac{|u|}{(3-u^2)^2}\right).$

If all else fails, the Weierstrass substitution should do things like this, but only if you can tolerate messy algebraic equations requiring numerical solutions. Let's try it:
$$
\begin{align}
& {} \qquad \int \frac{\Big(2t/(1+t^2)\Big)^3}{\Big(2t/(1+t^2)\Big)^3 - \Big((1-t^2)/(1+t^2)\Big)^3} \; \frac{2\;dt}{1+t^2} \\ \\ \\
& = \int \frac{(2t)^3}{(2t)^3 - (1-t^2)^3} \; \frac{2\;dt}{1+t^2} \\ \\ \\
& = \int\frac{8t^3}{t^6 - 3t^4 + 8t^3 + 3t^2 -1} \; \frac{2\;dt}{1+t^2}.
\end{align}
$$
At this point I think you'd have to result to numerical methods to factor the thing, but it looks as if the sixth-degree polynomial would be the product of two distinct first-degree factors and two irreducible quadratic factors. When you then find the partial fraction decomposition, there'd be the question of where you get $\text{constant}/\text{irreducible quadratic}$ (so you'd get an arctangent) and where you get $t/\text{irreducible quadratic}$ (so you'd get a logarithm).